Let I(x)=∫ x²(x sec²x+tanx)/(x tanx+1)²... | Definite Integration

JEE-Main 2023

08-04-2023_S1

Question

Let $I(x)=\int \frac{x^2\left(x \sec ^2 x+\tan x\right)}{(x \tan x+1)^2} d x$. If $I(0)=0$ the $I\left(\frac{\pi}{4}\right)$ is equal to

Select the correct option:

$\log _e \frac{(\pi+4)^2}{16}-\frac{\pi^2}{4(\pi+4)}$

$\log _e \frac{(\pi+4)^2}{16}+\frac{\pi^2}{4(\pi+4)}$

$\log _e \frac{(\pi+4)^2}{32}-\frac{\pi^2}{4(\pi+4)}$

$\log _e \frac{(\pi+4)^2}{32}+\frac{\pi^2}{4(\pi+4)}$

✓ Correct! Well done.

✗ Incorrect. Try again or view the solution.

Solution

Question Tags

JEE Main

Mathematics

Medium

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